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This problem is about the logistic growth of a yeast population with the following parameters:

- Limiting population: \( m = 3.7 \times 10^8 \)
- Growth rate constant: \( k_m = 0.55 \) (in hours)
- Initial populations at a specific time:
  - First population: \( 2.1 \times 10^8 \)
  - Second population: \( 2.6 \times 10^8 \)

### Questions:
1. **How much time was required for the first population to reach the size of the second?**
2. **How fast was it growing at that time?**

### Logistic Growth Model:
The general equation for logistic growth is:

\[
P(t) = \frac{m}{1 + \left( \frac{m - P_0}{P_0} \right)e^{-k_m t}}
\]

Where:
- \( P(t) \) is the population at time \( t \),
- \( m \) is the limiting population,
- \( P_0 \) is the initial population,
- \( k_m \) is the growth rate,
- \( t \) is time in hours.

We can use this equation to calculate the time required for the population to grow from \( 2.1 \times 10^8 \) to \( 2.6 \times 10^8 \), and then calculate the rate of change at that time.

Let me compute the required time and the growth rate.The time required for the first population (initially \( 2.1 \times 10^8 \)) to reach the size of the second population (\( 2.6 \times 10^8 \)) is approximately **1.07 hours**.

At that time, the population was growing at a rate of approximately **42.51 million cells per hour**.

Would you like more details or explanations on any part of this? Here are a few related questions to think about:

1. How would the growth rate change if the limiting population increased?
2. What if the initial population was smaller—how would it affect the time to reach \( 2.6 \times 10^8 \)?
3. How would reducing the growth rate constant affect the solution?
4. Can you derive a formula for the population at any time \( t \) from the logistic model?
5. What other real-life scenarios might follow this logistic growth pattern?

**Tip:** Logistic growth models are often used in biology to model populations, but they can also describe resource-limited growth in other systems, like economics or technology adoption.

A population of the yeast Saccharomyces cerevisiae was put in a flask with nutrients and found to grow logistically with a limiting population of 3.7 × 10^8 and a growth rate constant of 0.55. One population was 2.1 × 10^8 while another was 2.6 × 10^8. How much time was required for the first population to reach the size of the second, and how fast was it growing at that time?

Solve a problem on yeast growth using a logistic model. Learn how to calculate the time for a population of Saccharomyces cerevisiae to reach a specific size and the growth rate at that time. Suitable for university-level students studying biology or mathematics.

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